ALLEN KVPY -2020 / Srea-SB/SX KVPY – 2020 STREAM - SB/SX

KVPY-2020 / Stream-SB/SX

1

ALLEN

PART-I : MATHEMATICS1. Consider the following statements :

I. n n

nn

2 ( 2)lim

2®¥

+ - does not exist

II.n n

nn

3 ( 3)lim

4®¥

+ - does not exist

Then(A) I is true and II is false(B) I is false and II is true(C) I and II are true(D) neither I nor II is true

Ans. (A)

Sol. I.

nn

nn

2 2lim

2 2®¥

æ ö-æ ö+ç ÷ç ÷ç ÷è øè ø

( )( )n

nlim 1 1

®¥= + - does not exist

II.

n n

n

3 3lim 0 0 0

4 4®¥

æ ö-æ ö æ ö+ = + =ç ÷ç ÷ ç ÷ç ÷è ø è øè ø2. Consider a regular 10-gon with its vertices on

the unit circle. With one vertex fixed, drawstraight lines to the other 9 vertices. Call themL1, L2,...,L9 and denote their lengths byl1,l2,..., l9 respectively. Then the productl1,l2,..., l9 is

(A) 10 (B) 10 3

(C) 50

3(D) 20

Ans. (A)

Sol. a4

a5

a6

a7a8

a9

a

a2a3l3

l2

l11

Let

2i i10 5e e

pæ ö pç ÷è øa = =

KVPY – 2020 STREAM - SB/SXPAPER WITH SOLUTION

Now, z10 – 1 = (z – 1) (z – a)...(z – a9) ....(1)so, l

1 l

2...l

9 = |1 – a| |1 – a2|....|1 – a9|

= |(1 – a) (1 – a2) ... (1 – a9)|

10

z 1

z 1lim 10

z 1®

-= =

-

3. The value of the integral/2 2

x/2

sin xdx

1 e

p

-p +ò

is

(A) 6

p(B)

4

p

(C) 2

p(D)

2

2

p

Ans. (B)

Sol./2 2 2

x x0

sin x sin xI dx

1 e 1 e

p

-

æ ö= +ç ÷+ +è ø

ò/2

2

0

1sin x dx

2 2 4

p p p= = ´ =ò

4. Let ¡ be the set of all real numbers andf(x) = sin10x (cos8x + cos4x + cos2x + 1)for x Î ¡. Let S = {l Î ¡ | there exits a pointc Î (0, 2p) with ƒ'(c) = lf(c)}.Then(A) S = ¡(B) S = {0}(C) S = [0, 2p](D) S is a finite set having more than one

elementAns. (A)Sol. Let g(x) = ƒ(x)e–lx; x Î [0, 2p]

so, g(0) = g(2p) = 0 (as ƒ(0) = ƒ(2p) = 0)

Thus, $ c Î (0, 2p)

such that g'(c) = 0

Þ ƒ'(c) = lƒ(c) " l Î ¡

Þ S = ¡

ALLEN


2

ALLEN5. A person standing on the top of a building of

height 60 3 feet observed the top of a towerto lie at an elevation of 45°. That persondescended to the bottom of the building andfound that the top of the same tower is now atan angle of elevation of 60°. The height of thetower (in feet) is

(A) 30 (B) 30( 3 1)+

(C) 90( 3 1)+ (D) 150( 3 1)+Ans. (C)

Sol.+

=60 3 d 3

d d

d60°

45°

60 3Ö–60 3Ö–

Þ =-

60 3d3 1

Þ æ ö= +ç ÷-è ø

1h 60 3 13 1

( )´= = +

-60 3 90 3 1

3 1

6. Assume that 3.13 < p < 3.15. The integerclosest to the value of sin–1(sin 1 cos 4 + cos 1sin 4), where 1 and 4 appearing in sin and cosare given in radians, is :(A) –1 (B) 1(C) 3 (D) 5

Ans. (A)Sol. q = sin–1(sin5)

y=sin (sinx)–1

0 p 2p5

= –(2p – 5)

= 5 – 2p » –1.26 (as p » 3.13)

7. The maximum value of the functionƒ(x) = ex + x ln x on the interval 1 < x < 2 is(A) e2 + ln 2 + 1 (B) e2 + 2 ln 2

(C) /2e ln2 2

p p p+ (D) 3/2 3 3

e ln2 2

+

Ans. (B)

Sol. ƒ'(x) = ex + 1 + lnx > 0

(as x Î [1, 2])

Þ ƒ(x) increases in [1, 2]

Þ ƒmax

= ƒ(2) = e2 + 2ln2

8. Let A be a 2 × 2 matrix of the form A =é ùê úë û1 1

ba,

where a, b are integers and –50 < b < 50. Thenumber of such matrices A such that A–1, theinverse of A, exists and A–1 contains onlyinteger entries is

(A) 101 (B) 200

(C) 202 (D) 1012

Ans. (C)

Sol. |A| ¹ 0 Þ a – b ¹ 0

Þ a ¹ b ...(i)

Also,

T

1 1 11A

a b b a- -é ù= ê ú- -ë û

-é ù= ê ú- -ë û

1 b1

a b 1 a

Thus, a – b = 1 or –1 ...(ii)

So, required number of pairs (a, b) is

101 × 2 = 202

9. Let A = ( )ij 1 i, j 3a

£ £ be a 3 × 3 invertible matrix

where each aij is a real number. Denote the

inverse of the matrix A by A–1. If 3

ijj 1a 1

==å

for 1 < i < 3 , then

(A) sum of the diagonal entries of A is 1

(B) sum of each row of A–1 is 1

(C) sum of each row and each column of A–1

is 1

(D) sum of the diagonal entries of A–1 is 1

ALLEN


3

ALLENAns. (B)Sol. Sum of elements in each row of A is 1.

So,

1 1

A 1 1

1 1

é ù é ùê ú ê ú=ê ú ê úê ú ê úë û ë û

1 1

1 1

A A 1 A 1

1 1

- -

é ù é ùê ú ê úÞ =ê ú ê úê ú ê úë û ë û

1

1 1

1 A 1

1 1

-

é ù é ùê ú ê úÞ =ê ú ê úê ú ê úë û ë û

Þ sum of elements in each row of A–1 is 1.

10. Let x, y be real numbers such that x > 2y > 0and 2 log (x – 2y) = log x + log y.

Then the possible value (s) of x

y

(A) is 1 only(B) are 1 and 4(C) is 4 only(D) is 8 only

Ans. (C)Sol. log(x – 2y)2 = log(xy)

Þ (x – 2y)2 = xy

æ ö-Þ =ç ÷

è ø

2x 2y x

y y

æ ö æ öÞ - + =ç ÷ ç ÷

è ø è ø

2x x

5 4 0y y

Þ =x

1, 4y

Þ =x

4y

æ ö>ç ÷

è ø

xas 2

y

11. Let 2 2

2 2

x y1(b a)

a b+ = < , be an ellipse with major

axis AB and minor axis CD. Let F1 and F2 beits two foci, with A, F1, F2, B in that order onthe segment AB. Suppose ÐF1CB = 90°. Theeccentricity of the ellipse is

(A) 3 1

2

- (B) 1

3

(C) 5 1

2

- (D) 1

5

Ans. (C)

Sol. ´ = --b b

1ae a

C(0,b)

A(ae,0)

F2 F1B(–a,0)

Þ b2 = a2eÞ a2(1 – e2) = a2eÞ e2 + e – 1 = 0

- +Þ =

1 5e

2

12. Let A denote the set of all real numbers x suchthat x3 – [x]3 = (x – [x])3, where [x] is thegreatest integer less than or equal to x. Then(A) A is a discrete set of at least two points(B) A contains an interval, but is not an interval(C) A is an interval, but a proper subset of

(–¥, ¥)(D) A = (–¥, ¥)

Ans. (B)Sol. (x – [x]) (x2 + [x]2 + x[x])

= (x – [x]) (x2 + [x]2 – 2x[x])Þ (x – [x]) (3x[x]) = 0Þ x = 0 or [x] = 0 or x = [x]Þ x Î Z È [0, 1)

13. Define a sequence {sn} of real numbers byn

n 2k 0

1s for n 1

n k '=

= ³+

å

Then nnlim s

®¥

(A) does not exist(B) exists and lies in the interval (0,1)(C) exists and lies in the interval [1, 2)(D) exists and lies in the interval [2, ¥)

ALLEN


4

ALLENAns. (C)

Sol. Since, = = =

£ £+ + +

å å ån n n

2 2 2k 0 k 0 k 0

1 1 1

n n n k n 0

®¥ ®¥ ®¥Þ £ £

+n2 2n n n

n nlim lim S lim

n n n

®¥Þ £ £n

n1 lim S 1

®¥Þ £n

nlim S 1

14. Let ¡ be the set of all real numbers andƒ : ¡ ® ¡ be a continuous function. Suppose|ƒ(x) – ƒ(y)| ³ |x – y| for all real numbers xand y. Then(A) ƒ is one-one, but need not be onto(B) ƒ is onto, but need not be one-one(C) ƒ need not be either one-one or onto(D) ƒ is one-one and onto

Ans. (D)Sol. Let ƒ(x) = ƒ(y)

So, |ƒ(x) – ƒ(y)| ³ |x – y|Þ 0 ³ |x – y| Þ x – y = 0 Þ x = yÞ ƒ is one-oneSince, ƒ is continuousSo ƒ(0) is finiteNow, |ƒ(x) – ƒ(0)| ³ |x – 0|

Þ ®¥ ®¥

- ³x xlim ƒ(x) ƒ(0) lim x

Þ ®¥

= ¥xlim ƒ(x)

Þ ƒ is unboundedÞ ƒ is surjective

15. Let x

, x (0,1)f(x) sin x

1, x 0

ì Îï= íï =î

,

Consider the integral

In = 1/n

nx

0

n f(x) e dx-ò

Then ®¥ n

nlim I

(A) does not exist (B) exists and is 0(C) exists and is 1 (D) exists and is 1 – e–1

Ans. (B)Sol. ƒ(x) is an increasing function.

so, é öÎ ÷êë ø

1ƒ(x) 1,

sin1" x Î [0, 1)

Now,

- - -£ £ò ò ò

1/ n 1/ n 1/ nnx nx nx

0 0 0

nn e dx n ƒ(x)e dx e dx

sin1

Þ ( )®¥ ®¥

- -£ £nn n

1 11 1

e elim lim In sin1 n

Þ ®¥£ £nn

0 lim I 0

Þ ®¥

=nnlim I 0

16. The value of the integral

34 3 2019

1

((x 2) sin (x 2) (x 2) 1)dx- - + - +ò

is(A) 0 (B) 2(C) 4 (D) 5

Ans. (B)

Sol. ( ) ( ) ( )( )- - + - +ò3

4 20193

1

x 2 sin x 2 x 2 1 dx

x – 2 = t Þ dx = dt

( ) ]-- -

+ + = = =ò ò1 1

14 3 20191

1 1

t sin t t 1 dt dt t 2

17. In a regular 15-sided polygon with all itsdiagonals drawn, a diagonal is chosen atrandom. The probability that it is either ashortest diagonal nor a longest diagonal is

(A) 2

3(B)

5

6

(C) 8

9(D)

9

10

ALLEN


5

ALLENAns. (A)

Sol. Total diagonals = 15C2 – 15 = 90

Shortest diagonal = Diagonal connecting

(A1A

3, A

2A

4, ...)

= 15

A1A2A3

A4

A5

A6

A7A8 A9

A10

A11

A12

A13

A14

A15

longest diagonal = Diagonal connecting

(A1A

8, A

1A

9, ...)

= 15

Required probability = - -90 15 15

90

= =60 2

90 3

18. Let M = 230 – 215 + 1, and M2 be expressed inbase 2. The number of 1’s in this base 2representation of M2 is

(A) 29 (B) 30

(C) 59 (D) 60Ans. (B)

Sol. ( ){

n2

n times

2 100...0=

M2 = (260 – 246) + (230 – 216) + 231 + 1

{{ { {( )14 times 14 times46 times 31 times16 times 2

11...100...0 11...1000...0 100...0 1= + + +123

Number of 1's = 14 + 1 + 14 + 1 = 30

19. Let ABC be a triangle such that AB = 15 andAC = 9. The bisector of ÐBAC meets BC inD. If ÐACB = 2ÐABC, then BD is

(A) 8 (B) 9

(C) 10 (D) 12

Ans. (C)Sol. In DABC

A

B CD

915

2qq

5 : 3

= =q q q

15 9 BC

sin2 sin sin 3

=q q

15 9

sin2 sin Þ q =

5cos

6

=q q

9 BC

sin sin 3

Þ BC = 9[3 – 4sin2q] = 9[4cos2q – 1]

é ù= ´ -ê úë û

259 4 1

36

= 16

\ BD = 5

8BC = 10

20. The figure in the complex plane given by2 2 2 210zz 3(z z ) 4i(z z ) 0- + + - =

is(A) a straight line (B) a circle(C) a parabola (D) an ellipse

Ans. (A)

Sol. ( )( ) ( )( )( )- + - + + - =2

10zz 3 z z 2zz 4i z z z z 0

Let z = x + iy10(x2 + y2) – 3(4x2 – 2x2 – 2y2) + 4i(2x(2iy)) = 0Þ 4x2 + 16y2 – 16xy = 0Þ x2 – 4xy + 4y2 = 0Þ (x – 2y)2 = 0 Þ x = 2y\ straight line

PART-I : PHYSICS21. Students A, B and C measure the length of a

room using 25 m long measuring tape of leastcount (LC) 0.5 cm, meter-scale of LC 0.1 cmand a foot-scale of LC 0.05 cm, respectively. Ifthe specified length of the room is 9.5 m, thenwhich of the following students will report thelowest relative error in the measured length ?(A) Student A(B) Student B(C) Student C(D) Both, student B and C

ALLEN


6

ALLENAns. (A)Sol. Student A : Length of scale = 25 m

Least count = 0.5 cm = 0.005mStudent A can measure the length of 9.5m byusing the scale only once so there will be anerror of 0.005m in 9.5m

\ Relative error = =0.005

0.00059.5

Student B : Length of scale : 1m = 100 cm

Least count = 0.05 cmTo measure 9.5m, student B has to use this meterscale atleast 10 times

\ Relative error = ´ =0.05

10 0.005100

cm

Student C : Length of scale : 1 foot = 30.48cm

Least count = 0.05 cm

To measure 9.5m, student C has to use this scaleapproximately 31 times

\ Relative error = ´ =0.05

31 0.0530.48

cm

\ Relative error is least for Student A.22. Meena applies the front brakes while riding on

her bicycle along a flat road. The force that slowsher bicycle is provided by the

(A) front tyre(B) road(C) rear tyre

(D) brakesAns. (B)Sol. The frictional force on the tyres is an external

force and is being provided by the road.

Other options i.e. front tyre, rear tyre and brakescomprise the internal parts of bicycle thus forcesapplied by them will be internal only.

23. A proton and an antiproton come close to eachother in vacuum such that the distance betweenthem is 10 cm. Consider the potential energy tobe zero at infinity. The velocity at this distancewill be :(A) 1.17 m/s(B) 2.3 m/s

(C) 3.0 m/s(D) 23 m/s

Ans. (A)Sol. Applying mechanical energy conservation

Ki + Ui = Kf + Uf

( )( )æ ö+ = + +ç ÷è ø

1 22 2 k q q1 10 0 mv mv

2 2 r

|q1| = |q2| = e, q1 = +e

q2 = – e

and r = 0.1 m

\ =2

2 kemv

r

\( )

( )( )

-

-

´ ´ ´= =

´

29 1922

27

9 10 1.6 10kev

mr 1.67 10 0.1

v ; 1.17 m/s

24. A point particle is acted upon by a restoringforce –kx3. The time period of oscillation is Twhen the amplitude is A. The time period foran amplitude 2A will be :

(A) T

(B) T/2

(C) 2T

(D) 4T

Ans. (B)Sol. Given F = –kx3

- = - 3dUkx

dx

Þ = 41U kx

4

\ Energy of oscillations will be

æ ö= + = +ç ÷è ø

22 41 1 dx 1

E mv U m kx2 2 dt 4

.... (1)

If we pull =dx

0dt

in above equation, we will

get amplitude as = 44E

Ak

... (2)

ALLEN


7

ALLENAlso on rearranging equation (1), we get

-æ ö= ± -ç ÷è ø

1/24m k

dt dx 1 x2E 4E

Now, use = 44E

Ak

, to reduce above equation

as

--

æ öæ öç ÷= ± - ç ÷ç ÷è øè ø

1/2422m x

dt dx A 1k A

The time period can be found by integratingabove equation.

--

æ öæ öç ÷= - ç ÷ç ÷è øè øò

1/ 24A2

0

2m xT 4 dx A 1

k A

--

æ öæ öç ÷= - ×ç ÷ç ÷è øè øò

1/ 24A2

0

2m x4 A 1 dx

k A

Put = Þ =x

u dx AduA

\ T = 4 2m

kA–2 (A) ( )-

-ò1 1/ 24

0

du 1 u

( )-= 12mT 4 A I

k

Where ( )-= -ò

1 1/ 24

0

I 1 u du is a numerical value

So from above equation T µ A–1

\ = Þ =12

2

T 2A TT

T A 2

25. The output voltage (taken across the resistance)of a LCR series resonant circuit falls to half itspeak value at a frequency of 200 Hz and againreaches the same value at 800 Hz. Thebandwidth of this circuit is :

(A) 200 Hz

(B) 200 3 Hz

(C) 400 Hz

(D) 600 Hz

Ans. (B)

Sol. VOutput = VR

= irmsR

= ( )=

+ -

0 0

22L C

V R V R

Z R X X

For peak XL = XC Þ Vpeak = V0

For VOutput = 0V

2

( )=

+ -

0 0

22L C

V V R

2 R X X

R2 + (XL – XC)2 = 4R2

- = ±L CX X 3R

w - = ±w1

L 3RC

w w - =m2LC 3R C 1 0

± ± +w =

2 23RC 3R C 4LC

2LC

- + +w =

2 2

13RC 3R C 4LC

2LC = 200 × 2p

+ + +w =

2 2

23RC 3R C 4LC

2LC = 800 × 2p

w2 – w1 = 600 × 2p = R

3L

Bandwidth = p´

=R 2 600

L 3

D = = =p1 R 600

f 200 32 L 3

ALLEN


8

ALLEN26. A collimated beam of charged and uncharged

particles is directed towards a hole marked Pon a screen as shown below. If the electric andmagnetic fields as indicated below are turned on

PE

BO u t o fth e p lan e

®

®

(A) only particles with speed E/B will gothrough the hole P.

(B) only charged particles with speed E/B andneutral particles will go through P.

(C) only neutral particles will go through P.

(D) only positively charged particles with speedE/B and neutral particles will go through P.

Ans. (C)

Sol. For charged particles

v

qE+qvB

net force is in downward direction, so they won’tbe able to go through the hole P.

And uncharged particle don’t deviate so theywill be able to go through hole P.

27. An engine runs between a reservoir attemperature 200 K and a hot body which isinitially at temperature of 600 K. If the hot bodycools down to a temperature of 400 K in theprocess, then the maximum amount of work thatthe engine can do (while working in a cycle) is(the heat capacity of the hot body is 1 J/K)

(A) 200(1 – ln 2) J

(B) 200(1 – ln 3/2) J

(C) 200(1 + ln 3/2) J

(D) 200 J

Ans. (B)

Sol. h =in

W

Q

Þ W = hQin

= òQ Cdt

For maximum amount of work, efficiencyshould be maximum, means we have to assumecarnot engine.

\h = - = -2

1

T 2001 1

T T

æ ö\ = = - -ç ÷è øò ò

400

in600

200W nQ 1 CdT

T

[ ]= - - l400

600C T 200 nT

é ùæ ö= - - + ç ÷ê úè øë û

l3

W C 200 n 2002

C = 1 (Given)

æ ö\ = - ç ÷è ø

l3

W 200 200 n2

æ öæ ö= -ç ÷ç ÷è øè ø

l3

W 200 1 n2

28. The clocktower ("ghantaghar") of Dehradun isfamous for the sound of its bell, which can beheard, albeit faintly, upto the outskirts of thecity 8 km away. Let the intensity of this faintsound be 30 dB. The clock is situated 80 mhigh. The intensity at the base of the tower is :-(A)60 dB. (B) 70 dB.(C) 80 dB. (D) 90 dB.

Ans. (B)Sol. r2 = 80m, L2 = ?

r1 = 8000 m, L1 = 30 dB

æ ö= ç ÷

è ø10

0

IL 10 log

I

Intensity due to a point source, µ2

1I

r

æ ö æ ö- = -ç ÷ ç ÷

è ø è ø2 1

2 1 10 100 0

I IL L 10 log 10 log

I I

ALLEN


9

ALLEN

æ öæ ö- = = ç ÷ç ÷ ç ÷è ø è ø

22 1

2 1 10 10 21 2

I rL L 10 log 10 log

I r

L2 – 30 = 10 log10 (104) = 40

L2 = 70 dB

29. An initially uncharged capacitor C is being

charged by a battery of emf E through a

resistance R upto the instant when the capacitor

is charged to the potential E/2, the ratio of the

work done by the battery to the heat dissipated

by the resistor is given by :-

(A)2 : 1 (B) 3 : 1

(C) 4 : 3 (D) 4 : 1

Ans. (C)

Sol.

R C

E

-= t / RCEi e

R, ( )-= - t / RCQ CE 1 e

Capacitor is charged to E

2,

So =CE

Q2

( )-\ = - t / RCCECE 1 e

2

-= t / RC1e

2

t = RC ln 2

Work done by battery = (Qflown) (DV)

( )æ ö= =ç ÷è ø

2CE CEE

2 2

Heat dissipated = òlRC n2

2

0

i Rdt

-= ×òlRC n22

2t / RC

0

Ee dt

R

æ ö= ç ÷ç ÷

è ø

23 CE

4 2

= =æ öç ÷ç ÷è ø

2

2

Work done CE / 2 4

Heat dissipated 33 CE4 2

30. Consider a sphere of radius R with unform chargeddensity and total charge Q. The electrostaticpotential distribution inside the sphere is given by

c

0

Q(r) (a b(r / R) )

4 Rf = +

pe. Note that the

zero of potential is at infinity.The values of (a, b, c) are :-

(A)1 3

, ,12 2

æ ö-ç ÷è ø

(B) 3 1

, ,22 2

æ ö-ç ÷è ø

(C)1 1

, ,12 2

æ öç ÷è ø

(D) 1 1

, ,22 2

æ ö-ç ÷è ø

Ans. (B)Sol. Potential inside uniformly charged solid sphere

is given by

é ù= -ë û2 2

3

kQV 3R r

2R

é ù= -ê ú

ê úë û

2 2

2 2

kQ 3R r

R 2R 2R

é ùæ ö= ê - úç ÷pÎ è øê úë û

2

0

Q 3 1 r

4 R 2 2 R

Compare with given formula, i.e,

é ùæ öê + úç ÷pÎ è øê úë û

C

0

Q ra b

4 R R

= = - =3 1

a ,b ,c 22 2

ALLEN


10

ALLEN31. The efficiency of the cycle shown below in the

figure (consisting of one isobar, one adiabat andone isotherm) is 50%. The ratio, x, between thehighest and lowest temperatures attained in thiscycle obeys (the working substance is aideal gas) :-

isobar

adiabatisotherm

V

P

(A)x = ex–1 (B) x2 = ex–1

(C) 2x 1x e -= (D) 22 x 1x e -=Ans. (B)

Sol.

iso ba r

ad iab a tiso th e rm

P1

P2

T1

T1

T2

P

V

= g-P

P

C

C R

CP = gCP – gRgR Þ (g – 1)CP

g=

g - P

RC

1

h =Work done

Heat sup plied

h =

æ öD - ç ÷

è ø =D

l1

P2

P

PnC T nRT n

P 1

nC T 2

nCPDT = 2nRTln1

2

P

P

-g gæ ö æ ö

=ç ÷ ç ÷è ø è ø

1

1 1

2 2

P T

P T

( )g-

g - 2 1

RT T

1 = 2RTT1ln

g-gæ ö

ç ÷è ø

11

2

T

T

T2 – T1 = 2T1lnæ öç ÷è ø

2

1

T

T

x – 1 = ln(x2)x2 = ex – 1

Option (B)32. A right-angle isosceles prism is held on the

surface of a liquid composed of miscible solventsA and B of refractive index nA = 1.50 and nB =1.30, respectively. The refractive index of prismis np = 1.5 and that of the liquid is given by nL= CAnA + (1 – CA)nB, where CA is the percentageof solvent A in the liquid :-

n2

Liquidoutlet

Liquidinlet

qi

n = 1

If qC is the critical angle at prism-liquid interface,the plot which best represents the variation ofthe critical angle with the precentage of solventis:

(A)

90°80°70°60°50°

0.5 CA1

qC

0

(B)

90°80°70°60°50°

0.5 CA1

qC

0

(C)

90°80°70°60°50°

0.5 CA1

qC

0

(D)

90°80°70°60°50°

0.5 CA1

qC

0

Ans. (A)Sol. nPsinqC = nLsin90°

- æ öq = ç ÷

è ø1 L

CP

nsin

n

( )- æ ö+ -q = ç ÷

è ø

A A A B1C

C n 1 C nsin

1.5

® Graph between qC and CA will be curve ofsin–1,

ALLEN


11

ALLENCheck for CA = 0.5, to find most appropriategraph

( ) ( )- æ ö+q = ç ÷

è ø

1C

0.5 1.5 0.5 1.3sin

1.5

- æ öq = °ç ÷è ø

;1

C

14sin 69

15

\ Correct option is (A)33. Instead of angular momentum quantization a

sudent posits that energy is quantized asE = –E0/n (E0>0) and n is a positive integer.Which of the following options is correct?

(A)The radius of the electron orbit is r n.µ

(B) The speed of the electron is v n.µ

(C) The angular speed of the electron is w µ 1/n

(D)The angular momentum of the electron is

n.µ

Ans. (D)

Sol. =2

e

mvF

r Þ

( )( )=

2

2

k Ze emv

r r

=2

21 KZemv

2 2r.....(i) (Kinetic energy)

Potential energy = ( )( )-

=1 2K Ze eKq q

r r....(ii)

Total energy = KE + PE = - = -2

0EKZe

2r n

\ r µ n

As kinetic energy = 2KZe

2rÞ KE µ

1

n

or µ2 1v

n Þ µ2 1

vn

L = mvrL µ vr

( )µ µ1

L n nn

Þ µL n

34. A monochromatic beam of light is incident atthe interface of two materials of refracive indexn1 and n2 as shown. If n1 > n2 and qc is thecritical angle then which of the followingstatements is NOT true?

q2

q3q1

n1

n2

(A) q1 = q3 for all values of q1.(B) cosq2 is imaginary for q1>qc.(C) cosq2=0 for q1 = qc.(D)cosq3 is imaginary for q1 = qc.KVPY Ans. (D)ALLEN Ans. (B,C,D)

Sol. n1 > n2

this means light is going from rarer to densermedium.So q2 will always be less than q1

n2sinq1 = n1sinq2

So cos(q2) will never be imaginary and also q2

can’t be 90°.In question incorrect options are asked.\ (B,C,D)

35. The intensity of light from a continuouslyemitting laser source operating at 638 nmwavelength is modulated at 1 GHz. Themodulation is done by momentarily cutting theintensity off with a frequency of 1 GHz. Whatis the farthest distance apart two detectors canbe placed in the line of the laser light, so thatthey can see the portions of the same pulsesimultaneously? (Consider the speed of light inair 3 ×108m/s) :-(A)30 µm (B) 30 cm(C) 3m (D) 30 m

Ans. (B)

Sol. Time period between two flashes = 1

fDistance travelled by laser in this interval

´= = = =

8

9

c 3 100.3m 30cm

f 10

ALLEN


12

ALLENSo this is the maximum distance between twodetectors, so that they can see the same pulsesimultaneously.

36. A conducting rod, with a resistor of resistanceR, is pulled with constant speed v on a smoothconducting rail as shown in figure. A constant

magnetic field Br

is directed into the page.

If the speed of the bar is doubled, by what factordoes the rate of heat dissipation across theresistance R change ?

BR

V

(A) 0 (B) 2(C) 2 (D) 4

Ans (D)Sol. Emf = VBL

=VBL

IR

Heat = I2R = 2 2 2V B L

R

Given V1 = 2V

So =1H

4H

37. The time period of a body undergoing simpleharmonic motion is given by T = paDbSc, wherep is the pressure, D is density and S is surfacetension. The values of a, b and c respectivelyare

(A) 1, 1

2,

3

2(B)

3

2,

1

2- , 1

(C) 1, 1

2- ,

3

2(D)

3

2- ,

2

1, 1

Ans. (D)

Sol. [T’] = [maL–aT–2a mbL–3bmcT–2c][T’] = [ma+b+c L–a–3b T–2a–2c]a + b + c = 0–a – 3b = 0–2a – 2c = 1On solving

-=

3a

2, =

1b

2, c = 1

38. Consider the following statements regarding thereal images formed with a converging lens.(I) Real images can be seen only if the image isprojected onto the screen(II) The real image can be seen only from thesame side of the lens as that on which the objectis positioned.(III) Real images produced by converging lensesare not only laterally but also longitudinallyinverted as with mirrors.Which of the above statement/statements is/areincorrect ?(A) Only I and III (B) All three(C) None (D) Only II

Ans. (B)Sol. Theoretical ® B39. A zinc ball of radius, R = 1 cm charged to a

potential –0.5 V. The ball is illuminated by amonochromatic ultraviolet (UV) light with awavelength 290 nm. The photoelectric thresholdfor zinc is 332 nm. The potential of ball after aprolonged exposure to the UV is(A) –0.5 V (B) 0 V(C) 0.54 V (D) 0.79 V

Ans. (C)

Sol. f = =12431

3.74 eV3320

e = = =lhc 12431

4.28 eV2900

(KE)max = 4.28 – 3.74= 0.54 eVInitially sphere is negatively charged so e– willgo easily then potential becomes O. After thatas e– will leave the potential will increase till itreaches the stopping potential value, V0

eV0 = 0.54 eVV0 = 0.54 V

ALLEN


13

ALLEN40. A source simultaneously emitting light at two

wavelengths 400 nm and 800 nm is used in theYoung's double slit experiment. If the intensityof light at the slit for each wavelength is I0,then the maximum intensity that can be observedat any point on the screen is(A) I0 (B) 2I0

(C) 4I0 (D) 8I0

Ans. (D)Sol. At central maxima

Due to 400 nm = 4I0

Due to 800 nm = 4T0

Total Intensity = 8 I0

PART-I : CHEMISTRY41. The stability of

CMe2 CMe2

Me2N

CMe2

Me

Me

I II III

follows the order :

(A) I > II > III (B) II > I > III

(C) II > III > I (D) III > II > I

Ans. (B)

Sol.

Me

(I)

CMe

d.l.

–N

(II)

Me Me +M

MeC

Med.l.

Me Me

MeC

Meloc. due to

S.I.Reffect

(III)

stability : II > I > III

Note : In III carbocation is localised due to S.I.R.

effect

42. Among the following, the biodegradablepolymer is :

(A) polylactic acid

(B) polyvinyl chloride

(C) bakelite

(D) teflon

Ans.(A)Sol. Polyacetic acid is biodegradable polymer.

43. Among the following,

II

O

H

MeO O

H

O

Me

O

HO

HMe

I III

IV V

the compounds which can be reduced withformaldehyde and conc. aq. KOH, are :

(A) only II and V

(B) only I and V

(C) only II and III

(D) only I, II and IV

Ans. (A)Sol. Aldehyde without a-H give Cannizaro

reaction.In Cannizaro reaction alcohol and carboxylicacid salt is formed.

H – C – H +

O

(II)

Me – OC – H

O

50%KOH¾¾¾® H–COO–

CH – OH2

Me

+

O

ALLEN


14

ALLEN

H – C – H +

O

(V)

H

O

50%KOH¾¾¾® HCOO–

CH – OH2

+

44. An organic compound that is commonly usedfor sanitizing surfaces is :(A) acetylsalicylic acid(B) chloramphenicol(C) aspartame(D) cetyltrimethyl ammonium bromide

Ans. (D)Sol. Cetyltrimethyl ammonium bromide is used for

sanitizing agent.

45. The rates of reaction of NaOH with :

FNO2

I

F

NO2II

NO2

IIIF

follow the order :-

(A) II > I > III (B) II > III > I

(C) I > III > II (D) III > II > I

Ans. (C)

Sol.

F

(I)

NO2

F

NO2

(II)

F

NO2

(III)

I and II can react with NaOH but II do not react

at room temperature. I and III give reaction

because at O and P position electrone

withdrawing group is present.

46. The most suitable reagent for the conversion of

2-phenylpropanamide into 1-phenylethylamine

is :-

(A) H2,Pd/C (B) Br2,NaOH

(C) LiAlH4,Et2O (D) NaBH4,MeOH

Ans. (B)

Sol. CH – CH – C – NH3 2

Ph

O2Br NaOH+¾¾¾¾® CH – CH – NH3 2

Ph47. The compound X in the following reaction

scheme :

H C3

CO H2acid

hydrolysis X [H]reduction H C3 NH2

is :-

(A) acetonitrile

(B) methyl isocyanide

(C) acetaldehyde

(D) nitromethane

Ans. (A)

Sol. CH – C N3 ºH O (Hydrolysis)3

+

CH – COOH3

Reduction CH – CH – NH3 2 2

CH3 – CN is common name Þ Acetonitrile

48. A nucleus X captures a b particle and then emits

a neutron and g ray to form Y. X and Y are :-

(A) isomorphs (B) isotopes

(C) isobars (D) isotones

Ans. (D)

Sol. Q 0 QP –1 P –1X Y+ b ®

X and Y has same mass number, hence they are

isotones.

49. The boiling point (in °C) of 0.1 molal aqueous

solution of CuSO4.5H2O at 1 bar is closest to:

[Given: Ebullioscopic (molal boiling point elevation)

constant of water, Kb = 0.512 K Kg mol–1] :-

(A) 100.36 (B) 99.64

(C) 100.10 (D) 99.90

Ans. (C)

Sol. 2H O 2 24 2 4CuSO 5H O Cu SOÅ× ¾¾¾® + 1

i = 2

DTb = i.Kb . m = 2 × 0.512 × 0.1 = 0.1024

0b b bT T T 100 0.1024 100.10¢ = + D = + =

ALLEN


15

ALLEN50. A weak acid is titrated with a weak base.

Consider the following statements regarding thepH of the solution at the equivalence point :(i) pH depends on the concentration of acid and

base.(ii) pH is independent of the concentration of

acid and base

(iii) pH depends on the pKa of acid and pKb ofbase.

(iv) pH is independent of the pKa of acid andpKb of base.

The correct statements are :-

(A) only (i) and (iii) (B) only (i) and (iv)

(C) only (ii) and (iii) (D) only (ii) and (iv)

Ans. (C)

Sol. For salts of weak acid and weak base

pH = 7 + 12

(pKa – pKb)

pH is independent of concentration of acid and

base

51. Products are favoured in a chemical reaction

taking place at a constant temperature and

pressure. Consider the following statements :

(i) The change in Gibbs energy for the reaction

is negative.

(ii) The total change in Gibbs energy for the

reaction and the surroundings is negative.

(iii) The change in entropy for the reaction is

positive.

(iv) The total change in entropy for the reaction

and the surroundings is positive.

The statements which are ALWAYS true are :-

(A) only (i) and (iii) (B) only (i) and (iv)

(C) only (ii) and (iv) (D) only (ii) and (iii)

Ans. (B)Since products are formed in the chemical

reaction taking place at constant temperature

and pressure, we can say that the reaction is

spontaneous.

Hence, DGreaction < 0

DStotal > 0

52. A mixture of toluene and benzene forms a nearlyideal solution. Assume P°B and P°T to be the vaporpressures of pure benzene and toluene,respectively. The slope of the line obtained byplotting the total vapor pressure to the molefraction of benzene is :-(A) P°B – P°T (B) P°T – P°B(C) P°B + P°T (D) (P°B + P°T)/2

Ans. (A)

Sol. 0 0 0 0total B B T T B B B TP (P ) (P ) (P ) (1 – )(P )= c + c =c + c

Comparing it with y = mx + c

total BPy x

c= ( )0 0

B T

m

P – PE55555F +

0Tc

PE5F

53. Upon dipping a copper rod, the aqueous solutionof the salt that can turn blue is :-(A) Ca(NO3)2 (B) Mg(NO3)2

(C) Zn(NO3)2 (D) AgNO3

Ans. (D)

Sol. 2AgNO3 + Cu ¾¾® Ag + Cu(NO3)2

Metal can reduce that metal cation which is

placed below it in reactivity series.

54. Treatment of alkaline KMnO4 solution with KIsolution oxidizes iodide to :-(A) I2 (B) IO–

4

(C) IO–3 (D) IO–

2

Ans. (C)Sol. KI + 2KMnO4 + H2O ® KIO3 + 2MnO2

+ 2KOH

55. If an extra electron is added to the hypotheticalmolecule C2, this extra electron will occupy themolecular orbital :-(A) p2P

* (B) p2P

(C) s2P* (D) s2P

Ans. (D)Sol. Configuration of C2

2 2 2 2 2 21s 1s 2s 2s 2px 2py 2pz* *s s s s p = p s

an extra electron added to the s2p of the above

configuration.

56. Among the following, the square planargeometry is exhibited by :

(A) CdCl42– (B) Zn(CN)4

2–

(C) PdCl42– (D) Cu(CN)4

3–


16

ALLENAns. (C)Sol. Complex Shape Hybridisation.

[CdCl4]2– Tetrahedral sp3

[Zn(CN)4]2– Tetrahedral sp3

[PdCl4]2– Square Planar dsp2

[Cu(CN)4]3– Tetrahedral sp2

57. The correct pair of orbitals involved inp-bonding between metal and CO in metalcarbonyl complexes is :

(A) metal dxy and carbonyl px*

(B) metal dxy and carbonyl px

(C) metal 2 2x yd

- and carbonyl px*

(D) metal 2 2x yd

- and carbonyl px

Ans. (A)Sol.

+—

+

M C Od

o rb italx y

—

+

+

p* M O

+

–

–

58. The magnetic moment (in mB) of

[Ni(dimethylglyxoimate)2] complex is closestto:

(A) 5.37 (B) 0.00

(C) 1.73 (D) 2.25

Ans. (B)

Sol. [Ni(dmg)2]

Ni2+

dmg–1 SFL

No unpaired electron present in splited ‘d’ obirtal

\ dimagnetic (mB = 0).

59. A compound is formed by two elements M andN. Element N forms hexagonal closed pack arraywith 2/3 of the octahedral holes occupied by M.The formula of the compound is :

(A) M4N3 (B) M2N3

(C) M3N2 (D) M3N4

Ans. (B)

Sol. Number of atoms of element ‘N’ per unit cell = 6

Number of atoms of element M per unit cell

= 23

(Number of octahedral voids per unit cell)

= 23

× 6 = 4

M : N = 4 : 6 = 2 : 3

Formula is M2N3

60. If the velocity of the revolving electron of He+

in the first orbit (n = 1) is v, the velocity of theelectron in the second orbit is :

(A) v (B) 0.5 v(C) 2v (D) 0.25 v

(D) This reaction involves an induced fitmechanism in hexokinase

Ans. (B)

Sol. For single electron species n1n

n µ

2 1

1 2

n 1n 2

n= =

n

2 11 1 0.52 2 2

nn = n = n = = n

PART-I : BIOLOGY61. Species with high fecundity, high growth rates,

and small body sizes are typically(A) endangered species

(B) keystone species

(C) K-selected species

(D) r-selected species

Ans. (D)62. When RNase enzyme is denatured by adding

urea, which ONE of the following combinationsof bonds would be disrupted?

(A) Ionic and disulphide bonds

(B) Ionic and hydrogen bonds

(C) Hydrogen and peptide bonds

(D) Peptide and disulphide bonds

Ans. (B)

ALLEN


17

ALLEN63. The function of aposematic colouration is to

(A) attract mates.

(B) camouflage.

(C) scare off competitors.

(D) warn predators.

Ans. (D)

64. Maize and rice genomes have diploidchromosome number of 20 and 24, respectively.In the absence of crossing over and mutations,which ONE of the following is CORRECTabout the genetic variation among theiroffspring?

(A) maize < rice

(B) maize = rice > 0

(C) maize = rice = 0

(D) maize > rice

Ans. (D)

65. The exponent z of the species-area curvemeasured at continental scales is

(A) smaller than the value of z at regional scales.

(B) equal to the value of z at regional scales.

(C) greater than the value of z at regional scales.

(D) unrelated to the value of z at regional scales.

Ans. (C)

66. The pH of an aqueous solution of 10–8 M HClis

(A) 6.0

(B) between 6.9 – 7.0

(C) between 7.0 – 7.1

(D) 8.0

Ans. (B)

67. Which ONE of the following can NOT causeeutrophication of lakes?

(A) Introduction of invasive floating plants

(B) Discharge of fertilizer-rich agriculturalwaste

(C) Natural ageing of lakes

(D) discharge of industrial waste

Ans. (D)

68. Which ONE of the following polymerasestranscribes 5S rRNA?

(A) RNA Pol I (B) RNA Pol III

(C) RNA Pol II (D) RNA Pol IV

Ans. (B)

69. Which ONE of the following statements aboutrennin is CORRECT?

(A) It is secreted by adrenal glands.

(B) It converts angiotensinogen to angiotensin.

(C) It is secreted by peptic cells of gastric glandsinto the stomach.

(D) It is a hormone.

Ans. (C)

70. When one goes from a brightly lit area to adimly lit room our eyes adjusts slowly, therebyregaining the clarity of vision. Which ONE ofthe following explains this process?

(A) Regeneration of rhodopsin in the rod cells(B) Bleaching of rhodopsin

(C) Constriction of the pupil

(D) Increase in the number of rod cells

Ans. (A)

71. In a diploid population at Hardy-Weinbergequilibrium, consider a locus a locus with twoalleles. The frequencies of these two alleles aredenoted by p and q, respectively. Heterozygosityin this population is maximum at

(A) p = 0.25, q = 0.75

(B) p = 0.4, q = 0.6

(C) p = 0.6, q = 0.4

(D) p = 0.5, q = 0.5

Ans. (D)

72. An enzyme with optimal acitivity at pH 2.0 and37°C is most likely to be:

(A) lysozyme from hen egg white

(B) trypsin from cattle

(C) DNA polymerase from Thermus aquaticus

(D)pepsin from humans

Ans. (D)

ALLEN


18

ALLEN73. While adjusting to varying environmental

temperature, plants incorporate in their plasmamembrane(A)more saturated fatty acids in cold and more

unsaturated fatty acids in hot environment.(B) more unsaturated fatty acids in cold and more

saturated fatty acids in hot environment.(C) more saturated fatty acids in both cold and

hot environment.(D)more unsaturated fatty acids in both cold and

hot environment.

Ans. (B)74. Which ONE of the following terms is NOT used

while describing human vertebra?

(A)Lumbar(B) Sacral(C) Thoracic(D) Tarsal

Ans. (D)75. Assume a population that has reached herd

immunity for an infectious disease. If an infectedindividual is introduced to this population, whichof the following is most likely to occur?

(A)The infection will spread exponentiallyacross the population.

(B) The infection will spread linearly across thepopulation.

(C) A few individuals may get infected, but theinfection will not spread across the population.

(D)No other individual will be infected by thedisease.

Ans. (C)76. Match the type of cells in Column I with the

organs they are part of, listed in Column II:

Column I Column IIP. Chondroblast i. Bone

Q. Osteoclast ii. Brain

R. Microglia iii. Cartilage

S. Pneumocyte iv. Lung

Choose the CORRECT combination.

(A)P-iii, Q-i, R-ii, S-iv

(B) P-ii, Q-i, R-iii, S-iv

(C) P-iv, Q-iii, R-ii, S-i

(D) P-iii, Q-ii, R-iv, S-i

Ans. (A)

77. A bacterial culture was started with an inoculumof 10 cells. What will be the number of cellsat the end of 10 cycles of division, assumingthat every progeny cell undergoes division ineach cycle?(A)100 (B) 1024(C) 2048 (D) 10240

Ans. (D)78. The following family tree traces the occurrence

of a rare genetic disease. The filled symbolssignify the individuals with the disease, whereasthe open symbols signify healthy individuals.

Based on this information, the disease is mostlikely to be

(A)autosomal, dominant

(B) autosomal, recessive

(C) X-linked, recessive

(D)X-linked, dominant

Ans. (B)

79. Which ONE of the following statements isCORRECT about the mechanism of action ofpenicillin?

(A) It inhibits transcription

(B) It hydrolyses cell wall

(C) It inhibits cell wall biosynthesis

(D) It inhibits translation

Ans. (C)

80. Leaf extract from an infected plant was passedthrough a filter with a pore size of 0.05 mmdiameter. The infectious agent was detected inthe filtrate. Which ONE of the following is thelikely infectious agent?

(A) Bacteria (B) Virus

(C) Nematode (D) Fungus

Ans. (B)

ALLEN


19

ALLENPART-II : MATHEMATICS

81. Let200 n

n

n 101 k 101

1a 2

k!= =

= å åand

b = 201 n200

n 101

2 2

n!=

-å

Then a

b is

(A)1 (B) 3

2

(C) 2 (D) 5

2Ans. (A)

Sol.= =

= å å200 n

n

n 101 k 101

1a 2

k!

æ ö æ ö= + + + + + +ç ÷ ç ÷è ø è ø

101102 1032 1 1 1 1 1

2 2 ...101! 101! 102! 101! 102! 103!

æ ö+ + + +ç ÷è ø

200 1 1 12 ...

101! 102! 200!

+ + + ++ + +

101 200 102 200 2002 ... 2 2 ... 2 2...

101! 102! 200!

( ) ( )- -= + + +

101 100 102 99 2002 2 1 2 2 1 2...

101! 102! 200!

æ ö æ ö æ ö= - + - + + -ç ÷ ç ÷ ç ÷

è ø è ø è ø

201 101 201 102 201 2002 2 2 2 2 2...

101! 101! 102! 102! 200! 200!

=

-= =å

201 n200

n 101

2 2b

n!

\ =a

1b

82. Let a, b, c be non-zero real roots of the equationx3 + ax2 + bx + c = 0. Then

(A) there are infinitely many such triples a, b, c(B) there is exactly one such triple a, b, c(C) there are exactly two such triples a, b, c(D) there are exactly three such triples a, b, c

Ans. (C)

Sol. x3 + ax2 + bx + c = 0 = (x – a)(x – b)(x – c)

a + b + c = –a

Þ 2a + b + c = 0 ...(i)

ab + bc + ca = b ...(ii)

abc = –c Þ ab = –1 [Q c ¹ 0] ...(iii)

Also a is a root of equation

Þ 2a3 + ab + c = 0 Þ 2a3 – 1 + c = 0

Þ c = 1 – 2a3

from (1)

2a2 + ab + ac = 0

2a2 – 1 + a(1 – 2a3) = 0

2a2 – 2a4 + a – 1 = 0

2a2(1 – a)(1 + a) + (a – 1) = 0

Þ (1 – a)[2a2(a + 1) – 1] = 0

Þ a = 1 or 2a3 + 2a2 – 1 = 0

when a = 1, b = -1

a = –1 and c = 1 – 2a3 = –1

when 2a3 + 2a2 – 1 = 0

There will be only one real solution of

ƒ(x) = 2x3 + 2x2 – 1 = 0

as ƒ'(x) = 6x2 + 4x = 0 Þ x = 0, -2

3

-æ ö <ç ÷è ø

2ƒ(0).ƒ 0

3

\ corresponding to this real value of a one tripletis possible

\ Exactly two triplets (a, b, c) are possible

83. Let ƒ(x) = sinx + (x3 – 3x2 + 4x – 2)cosx forx Î (0, 1). Consider the following statements

I. ƒ has a zero in (0, 1)

II. ƒ is monotone in (0, 1)

Then

(A) I and II are true

(B) I is true and II are false

(C) I is false and II are true

(D) I and II are false

ALLEN


20

ALLENAns. (A)Sol. ƒ(x) = sinx + (x3 – 3x2 + 4x – 2) cosx,

x Î (0,1)

ƒ(0) = –2 > 0

ƒ(1) = sin1 < 0

Q ƒ(0). ƒ(1) < 0 Þ ƒ(x) has a zero in (0,1)

Now,

ƒ(x) = sinx + [(x – 1)3 + (x – 1)] cosx

Þ ƒ'(x) = (3(x – 1)2 + 2)cosx – sinx[(x–1)3 + (x– 1)]= [3(x – 1)2 + 2] cosx + [(1 – x)3 + (1 – x)]sinx> 0 " x Î (0,1)Þ ƒ(x) is monotone in (0,1)

84. Let A be a set consisting of 10 elements. Thenumber of non-empty relations from A to A thatare reflexive but not symmetric is(A) 289 – 1 (B) 289 – 245

(C) 245 – 1 (D) 290 – 245

Ans. (D)Sol. n(A × A) = 100

number of (a,a) type pairs is 10number of (a,b) and (b,a) type pair of pairs is45 (a ¹ b)so, required number of relations is290 – 245

85. In a triangle ABC, the angle bisector BD of ÐBintersects AC in D. Suppose BC = 2, CD = 1

and BD = 3

2. The perimeter of the triangle

ABC is

(A) 17

2(B)

15

2

(C) 17

4(D)

15

4Ans. (B)

Sol.

B 2 C

D

A

3/ 2Ö

2ac B 3cos

a c 2 2=

+

94 14c 32

32 c 22 22

é ù+ -ê ú

Þ =ê ú+ ê ú´ ´

ê úë û

Þ c = 3

Now, c AD 3

ADa DC 2

= Þ =

5b

2Þ =

Þ Perimeter 15

2=

86. Let ¥ be the set of natural numbers.

For n Î ¥, define p

=+ò

2n

n 2n 2n0

xsin (x)I dx

sin (x) cos (x).

Then for m, n Î ¥

(A) Im < In for all m < n

(B) Im > In for all m < n

(C) Im = In for all m ¹ n

(D) Im < In for some m < n and Im > In for somem < n

Ans. (C)

Sol.( ) 2n2n

n 2n 2n 2n 2n0

x sin x1 xsin xI dx

2 sin x cos x sin x cos x

p æ öp -= +ç ÷ç ÷+ +è ø

ò

2n

2n 2n0

sin xdx

2 sin x cos x

pp=

+ò

/2 2n

2n 2n0

sin xdx2

2 sin x cos x

pp= ´

+ò

/2 2n 2n

2n 2n0

sin x cos xdx

2 sin x cos x

pp +=

+ò

2

2 2 4

p p p= ´ =

Þ Im = I

n " m,n

ALLEN


21

ALLEN87. For q Î [0, p], let f(q) = sin (cos q) and

g(q) = cos (sin q). Let a = 0max

£q£pf(q), b =

0min£q£p

f(q), c = 0max

£q£pg(q) and d =

0min£q£p

g(q). The correct

inequalities satisfied by a, b, c, d are

(A) b < d < c < a

(B) d < b < a < c

(C) b < d < a < c

(D) b < a < d < c

Ans. (C)Sol. ƒ(q) = sin(cosq)

g(q) = cos(sinq)

ƒ'(q) = cos (cosq) (– sinq) < 0 " q Î [0,p]

\ ƒ(q) decreases monotonically

\ a = max ƒ(q) = ƒ(0) = sin1

b = min ƒ(q) = ƒ(p) = –sin1

g'(q) = – sin(sinq) cosq

– +p/2

g(q) = 1 ; g(p) = 1; g cos12

pæ ö =ç ÷è ø

\ c = max g(q) = 1

d = min g(q) = cos1

\ b < d < a < c

88. Six consecutive sides of an equiangular octagonare 6, 9, 8, 7, 10, 5 in that order. The integernearest to the sum of the remaining two sidesis

(A) 17 (B) 18

(C) 19 (D) 20

Ans. (B)

Sol.P

y/ 2Ö H

y/ 2Ö

9/ 2Ö

9/ 2Ö 7/ 2Ö R

7/ 2ÖE

10

F

5/ 2Ö

G 5/ 2Ö

A

6

B

Q DC

y 5

798

x

Let ABCDEFGH be the equiangular octagon asshown PQ = SR

y 9 5 76 10

2 2 2 2Þ + + = + +

y 3 4 2Þ = +Also : PS = QR

y 5 9 7x 8

2 2 2 2Þ + + = + +

x 4 4 2Þ = +

\ x + y = 7 8 2+ = 18.313

\ Nearest integer = 18.

89. The value of the integral

22 1

21 4

x 1 1dx

x 1 1 x

+ æ ö-ç ÷

+è ø +ò is

(A) 6 2

p(B)

12 2

p

(C) 8 2

p(D)

4 2

p

Ans. (B)

Sol. ( )22 1

212

x 1dx

1 1x x x x

x x

+ -

æ ö+ +ç ÷è ø

ò

2 1 2

21

11

x dx1 1

x x 2x x

+ -=

æ ö æ ö+ + -ç ÷ ç ÷è ø è ø

ò

Let + = q1

x 2 secx

æ ö- = q q qç ÷è ø2

11 dx 2 sec tan d

xp

p

q q qq qò

/ 3

/ 4

2 sec tan d

2 sec 2 tan

12 2

p=

ALLEN


22

ALLEN90. Let a = BC, b = CA, c = AB be the side lengths

of a triangle ABC, and m be the length of themedian through A. If a = 8, b – c = 2, m = 6,then the nearest integer to b is

(A) 7 (B) 8

(C) 9 (D) 10

Ans. (B)

Sol.2 2 2

2 2b 2c am

4

+ -=

Þ 144 + 64 = 2[b2 + (b – 2)2]

Þ 104 = 2b2 – 4b + 4

Þ b2 – 2b – 50 = 0

Þ (b – 1)2 = 51

Þ ( )b 1 51 8,9= + Î

PART-II : PHYSICS91. A camera filled with a polarizer is placed on a

mountain, in a manner to record only the

reflected image of the sun from the surface of

a sea as shown in the figure. If the sun rises at

6.00AM and sets at 6.00PM during the summer,

then at what time in the afternoon will the

recorded image have the lowest intensity,

assuming there are no clouds and intensity of

the sun at the sea surface is constant throughout

the day? (Refractive index of water = 1.33)

West East

Sea

Camera

(A) 12.32PM (B) 3.32PM

(C) 5.00PM (D) 6.00PM

Ans. (B)

Sol.i i

Camera will receive minimum intensity when.Light will incident at Brewsters’s angle.\ tan i = µ = 4/3Þ i = 53°

9 0 °5 3 °

P

B A6 P M 6 A M

time taken by sun to go from A to P

will be ´ °°

12 hr143

180= 9.53 hr = 9 hr 32 min.

\ time = 6 AM + 9 hr 32 min Þ 3:32 PM

92. Suppose a long rectangular loop o width w ismoving along the x-direction with its left armin magnetic field perpendicular to the plane ofthe loop (see figure). The resistance of the loopis zero and it has an inductance L. At time, t =0, its left arm passes the origin, O.

y

x w v0

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

If for t ³ 0, the current in the loop is I and thedistance of its left arm from the origin is x, thenI versus x garph will be

(A)

I

x

(B)

I

x

(C)

I

x

(D)

I

x

Ans. (B)

ALLEN


23

ALLEN

Sol. v

x = u t

- =di

VB L 0dt

l

=di

VB Ldt

l

Þ =di VB

dt L

l

= +ve slope

x = ut Þ =dx

Vdt

=di B

dx L

l= +ve slope

93. Imagine a world where free magnetic charges

exist. In this world, a circuit is made with a U

shape wire and a rod free to slide on it. A current

carried by free magnetic charges can flow in

the circuit. When the circuit is placed in a

uniform electric field, E perpendicular to the

plane of the circuit and the rod is pulled to the

right with a constant speed v, the "magnetic

EMF" in the current and the direction of the

corresponding current, arising because of

changing electric flux will be ( l is the length of

the rod and c is speed of light ).

(A) vEl clockwise

(B) vEL counterclockwise

(C) 2vElc

clockwise.

(D) 2vElc

counterclockwise

KVPY Ans. (D)ALLEN Ans. (C or D)

Sol.æ öfæ ö= m + eç ÷ç ÷

è øè øò E

0 0

dB.d I

dt

uurrlÑ

f=Ed

vEdt

l

\ = m eò 0 0B.d (vE )uurrl lÑ Þ 2

vE

C

l

Direction of electric field is not given in thequestion therefore both options are possible.

94. The box in the circuit below has two inputs

marked v+ and v– and a single output marked

V0. The output obeys

+10V if v+ > v–

V0 =

–10V if v+ < v–

R

v–

v+ R

RC

v0

The output V0 of this circuit a long time after it

is switched on is best represented by

(A)

v0

+10V

–10V

t

(B)

+10V

–10Vt

(C) +5V

–5Vt

(D)

+10V

–10Vt

Ans. (A)

ALLEN


24

ALLENSol. V0 can only have two values either +10 or –10

\ Only (A) is possible

95. A bottle has a thin nozzle on top. It is filledwith water, held horizontally at a height of 1mand squeezed slowly by hands so that the waterjet coming out of the nozzle hits the ground at adistance of 2m. If the area over which the handssqueeze it is 10 cm2, the force applied by handsis close to (take g = 10 m/s2 and density of water= 1000 kg/m3)

1m2m

(A) 20 N (B) 10 N (C) 5 N (D) 2.5 N

Ans. (B)

Sol. 1 2

Apply Bernoulli between point-1 and point-2.

+ r = + r2 21 1 2 2

1 1P V P V

2 2

P1 = Patm + Force by hand

Area

V1 tends to zero becomes area of point-2 is verysmall.

P2 = Patm

Patm + F/A = Patm + (1/2) rV22 Þ =2

2

2FV

PA

....(i)

From Kinematics.

= ´ 2

2(h)2 V

g

\ V22 = 20 ....(ii)

Using (i) & (ii) we get

=r2F

20A \ F = 10N

96. The circular wire in figure below encirclessolenoid in which the magnetic flux is increasingat a constant rate out of the plane of the page.

wire

path 1

path 2ab

The clockwise emf around the circular loop ise0. By definition a voltammeter measures thevoltage difference between the two points given

by b

b a aV V E.ds- = - ò . We assume that a and b

are infinitesimally close to each other. Thevalues of Vb – Va along the path 1 andVa – Vb along the path 2, respectively are(A) –e0, –e0

(B) –e0, 0(C) –e0, e0

(D) e0, e0

Ans. (B)Sol. Flux is increasing while coming out of plane

\ Induced electric field will be in clockwisedirection.

\ òb

a

E.dsuurr

will be +ve e0.

for path-1Vb – Va = –e0

In path-2 if we see a & b very close and Netemf in path = 0

97. A beam of neutrons performs circular motion

of radius, r = 1 m, under the influence of an

inhomogeneous magnetic field with

inhomogeneity extending over Dr = 0.01 m. The

speed of the neutrons is 54 m/s. The mass and

magnetic moment of the neutrons respectively

are 1.67 × 10–27 kg and 9.67 × 10–27 J/T. The

average variation of the magnetic field over Dr

is approximately.

(A) 0.5 T

(B) 1.0 T

(C) 5.0 T

(D) 10.0 TAns. (C)

ALLEN


25

ALLEN

Sol.¶

= =¶

2B mvF M

r r

D = D2mv

B rMr

= -

-

´ ´ ´´ ´

27 2

27

1.67 10 54 0.01

9.67 10 1 = 5.03 T

98. A student is jogging on a straight path with thespeed 5.4 km per hour. Perpendicular to thepath is kept a pipe with its opening 8 m fromthe road (see figure). Diameter of the pipe is 0.45m. At the other end of the pipe is a speakeremitting sound of 1280 Hz towards the openingof the pipes. As the student passes in front ofthe pipe, she hears the speaker sound for Tseconds. T is in the range (take speed of sound,320 m/s) :

5.4 km/h

8 m

(A) 6-12(B) 12-18(C) 3-6(D) 18-22

Ans. (A)

Sol.

D

y

q

bO

l = =320

0.25m1280

Using concept of diffraction of wavebsinq = 1.22l

q = ´25

sin 1.2245

= 0.678

tanq = 0.93

q =y

tan2D

y Þ 2Dtanq

time to cross this region = q2Dtan

speed

Þ´ ´2 8 0.93

9.9sec1.5

;

99. A solar cell is to be fabricated for efficientconversion of solar radiation to emf usingmaterial A. The solar cell is to be mechanicallyprotected with the help of a coating usingmaterial B. If the band gap energy of materialsA and B are EA and EB respectively, then whichof the following choices is optimum for betterperformance of the solar cell.(A) EA = 1.5 eV, EB = 5 eV(B) EA = 1.5 eV, EB = 1.5 eV(C) EA = 3 eV, EB = 1.5 eV(D) EA = 0.5 eV, EB = 5 eV

Ans. (A)Sol. Generally we want the electron to cross energy

gaps in material A.Not in material-B because its just covering.And Eg in the A should not be very smallotherwise there will be huge heat loss becauseof large difference in Eg and energy of incidentphoton

100. The "Kangri" is an earthen pot used to staywarm in Kashmir during the winter months.Assume that the "Kangri" is spherical and ofsurface area 7 × 10–2 m2. It contains 300 g ofa mixture of coal, wood and leaves withcalorific value of 30 kJ/g (and provides heatwith 10% efficiency). The surface temperatureof the 'Kangri' is 60°C and the room temperatureis 0°C. Then, a reasonable estimate for theduration t (in hours) that the 'kangri' heat willlast is (take the 'kangri' to be a black body) :(A) 8 (B) 10(C) 12 (D) 16

Ans. (B)

Sol. = s -4 40 s

dQeA [T T ]

dte = 1, A = 7 × 10–2, s = 5.67 × 10–8

T0 = 333 K, Ts = 273 K

dQ

dt = 26.75 Watt

total energy produced = ´ ´ ´31030 10 300

100Þ 9 × 105J

\´

=´

59 10time hrs

26.75 3600 = 9.35 hrs

ALLEN


26

ALLENPART-II : CHEMISTRY

101. An organic compound X with molecular formulaC11H14 gives an optically active compound onhydrogenation. Upon ozonolysis, X produces amixture of compounds – P and Q. CompoundP gives a yellow precipitate when treated withI2 and NaOH but does not reduce Tollen'sreagent. Compound Q does not give any yellowprecipitate with I2 and NaOH but gives Fehling'stest. The compound X is :

(A) Ph

(B) Ph

(C) Ph (D) Ph

Ans. (A)Sol.

H /N i2

[X ](A c tiv e )

P h

P h

H*

o zo n o ly s is

×

tol le

n's

to ll e n 'sP h– C – C H + C H – C H – C –H3 3 2 C H – C H – C O O3 2

–

(P ) (Q )

P h– C – O N a + C H I ( l) 3

N aO H + I 2

O

y e llo w p p t.

O

y el lo w p p t . is n o t fo rm e d

N a O H + I 2

O

102. The following transformation

O

CO H2

O

can be carried out in three steps. The reagentsrequired for these three steps in their correctorder, are :

(A) (i) NaBH4; (ii) PCl5; (iii) anh.AlCl3

(B) (i) SOCl2; (ii) anh. AlCl3; (iii) Zn(Hg)/HCl

(C) (i) Zn(Hg)/HCl; (ii) SOCl2; (iii) anh.AlCl3

(D) (i) conc.H2SO4; (ii) H2N-NH2·H2O;(iii) KOH, ethylene glycol, D

Ans. (C)

Sol.

C O H2

OZn (H g )

con c . H C lC = O

H O

SO C l2

O+

O

C lC

C = OA n h . A lC l 3

103. In the following reaction

¾¾¾¾¾¾®2(i) O , catalyst , heat(ii) aq. acid

XOH

+ Y(C6H10O)

X and Y, respectively are :

(A) and CHO

(B) and O

(C) and O

(D) and O

Ans. (D)Sol.

O /h2 v O – O –H

H O [H /H O ]3 2

+ +

O+

sh iftingO — O

H

H

+

O

O H

O O H

+

H O2

ALLEN


27

ALLEN104. A two dimensional solid is made by alternating

circles with radius a and b such that the sidesof the circles touch. The packing fraction isdefined as the ratio of the area under the circlesto the area under the rectangle with sides oflength x and y.

a b

x = 2a + 2b

The ratio r = b/a for which the packing fractionis minimized is closest to :(A) 0.41 (B) 1.0

(C) 0.50 (D) 0.32Ans. (A)

y=2a

x=2a+2b

Area of rectangle = xy= 2a(2a+2b)= 4a(a+b)

Area covered by circles = pa2 + pb2 = p(a2+b2)

Packing fraction (P.F.) = p +

+

2 2

2

(a b )4(a ab)

æ öp +ç ÷

è ø=æ ö+ç ÷è ø

22

2

2

ba 1

a

b4a 1

a

æ ö=ç ÷è ø

bPutting r

a

p +=

+

2(1 r )P.F.

4(1 r)

For minimum P.F., =d(P.F.)

0dr

é ùp + - +=ê ú+ë û

2

2

2r(1 r) (1 r )or 0

4 (1 r)

Þ r2 + 2r – 1 = 0

- + += = - =

2 4 4or r 2 1 0.414

2

Answer is option (A)

105. Consider a reaction that is first order in bothdirections

f

b

k

kA B��

Initially only A is present, and its concentrationis A0. Assume At and Aeq are the concentrationsof A at time "t" and at equilibrium, respectively.The time "t" at which At = (A0 + Aeq)/2 is :

(A) f b

3ln2t

(k k )

æ öç ÷è ø=+

(B) f b

3ln2t

(k k )

æ öç ÷è ø=-

(C) f b

ln 2t(k k )

=+

(D) f b

ln 2t(k k )

=-

Ans. (C)

A B�

t = 0t = t

t = teq

[A ]0

[A ]–x0

[A ]–x0 eq

0x

xeq

= At

= Aeq

Given at time t = t At = ( )+0 aqA A

2and xeq. = A0 – Aeq

Now, t = -

æ ö æ ö=ç ÷ ç ÷+ +è ø è ø

ll

e

f b e x f b

x1 n2n

k k x k k

106. The reaction

3 2CaCO (s) CaO(s) CO (g)+��

is in equilibrium in a closed vessel at 298 K.The partial pressure (in atm) of CO2(g) in thereaction vessel is closest to:

[Given : The change in Gibbs energies offormation at 298 K and 1 bar for

CaO(s) = –603.501 kJ mol–1

CO2(g) = – 394.389 kJ mol–1

CaCO3(s) = – 1128.79 kJ mol–1

Gas constant R = 8.314 J K–1 mol–1]

(A) 1.13 × 10–23 (B) 0.95

(C) 1.05 (D) 8.79 × 1023

Ans. (A)

ALLEN


28

ALLEN

Sol. CaCO3(s) �� CaO(s) + CO2(g)

DrG0 = DfG

0 (CaO) + DfG0 (CO2) – DfG

0 (CaCO3)

= – 603.501–394.389 + 1128.79=130.9 kJmol–1

DrG0 = – 2.303 RT log Kp

log Kp = 130.9 ×1000

= -22.94-2.303 × 298 × 8.314

Kp = antilog (–22.94) = 1.13 × 10–23

107. A container is divided into two compartmentsby a removable partition as shown below :

n , VT, PHe

1 1 n , VT, PNe

2 2

removable partition

In the first compartment, n1 moles of ideal gasHe is present in a volume V1. In the secondcompartment, n2 moles of ideal gas Ne is presentin a volume V2. The temperature and pressurein both the compartments are T and P,respectively. Assuming R is the gas constant, thetotal change in entropy upon removing thepartition when the gases mix irreversibly is :

(A) 1 21 2

1 2 1 2

V Vn R n n R nV V V V

++ +

l l

(B) 1 2 1 21 2

1 2

V V V Vn R n n R nV V+ +

+l l

(C) 1 11 2

2 2

n V(n n )R nn V

+ l

(D) 2 21 2

1 1

n V(n n )R nn V

+ l

Ans. (B)Sol. Entropy change DS = nCVln

2 2

1 1

T VnR nT V

æ ö æ ö+ç ÷ ç ÷

è ø è øl

Since temperature is constant throughoutprocess.

He : DS = n1Rln 1 2

1

V VV+æ ö

ç ÷è ø

Ne : DS = n2Rln 1 2

2

V VV+æ ö

ç ÷è ø

Total change in (DS) = n1Rln 1 2

1

V VV+æ ö

ç ÷è ø

+ n2Rln 1 2

2

V VV+æ ö

ç ÷è ø

108. Number of stereoisomers possible for theoctahedral complexes [Co(NH3)3Cl3] and[Ni(en)2Cl2], respectively, are:

[en = 1,2-ethylenediamine]

(A) 2 and 4

(B) 4 and 3

(C) 3 and 2

(D) 2 and 3

Ans. (D)Sol. [Co(NH3)3Cl3]

Co CoCl Cl

Cl NH3

NH3 ClH N3 H N3

H N3 H N3

Cl Cl

optically inactive optically inactivefac mer

GI = 2Total steroisomer = 2

Co NiCl NH2

Cl NH2

NH2 H N2

H N2Cl

H N2 ClNH2 H N2

d

H C2

H C2

CH2 CH2

CH2 CH2

CH2

CH2

l

cis

[N i(e n ) C l ]2 2

Ni

ClNH2

H N2

Cl

H C2

H C2

NH2

H N2

CH2

CH2

Trans (optically inactive)

G.I. = 2

Optical isomer = 3

stereo isomer = 3

ALLEN


29

ALLEN109. When a mixture of NaCl, K2Cr2O7 and conc.

H2SO4 is heated in a dry test tube, a red vapor(X) is evolved. This vapor (X) turns an aqueoussolution of NaOH yellow due to the formationof Y. X and Y, respectively, are:

(A) CrCl3 and Na2Cr2O7

(B) CrCl3 and Na2CrO4

(C) CrO2Cl2 and Na2CrO4

(D) Cr2(SO4)3 and Na2Cr2O7

Ans. (C)

Sol.4NaCl + K2Cr2O7 + 6H2SO4 ® 4NaHSO4 + 2KHSO4

+ 2CrO2Cl2(X) + 6H2O

CrO2Cl2 + 4NaOH ¾® Na2CrO4(Y) + 2NaCl + 2H2O

X = CrO2Cl2

Y = Na2CrO4

110. Sodium borohydride upon treatment with iodine

produces a Lewis acid (X), which on heating

with ammonia produces a cyclic compound (Y)

and a colorless gas (Z). X, Y and Z are :

(A) X = BH3 ; Y = BH3. NH3 ; Z = N2

(B) X = B2H6 ; Y = B3N3H6 ; Z = H2

(C) X = B2H6; Y = B6H6; Z = H2

(D) X = B2H6 ; Y = B3N3H6; Z = N2

Ans. (B)

Sol. 2Na[BH4] + I2 ¾® B2H6(X) + 2NaI + H2

3B2H6 + 6NH3 D¾¾® 2B3N3H6(Y) + 12H2(Z)

X = B2H6

Y = B3N3H6

Z = 12 H2

PART-II : BIOLOGY111. Which ONE of the following is the most likely

ratio of blood groups (A : B : AB) among theprogeny from heterozygous parents with B andAB blood groups ?(A) 0.5 : 0.25 : 0.25(B) 0.25 : 0.25 : 0.5(C) 0.25 : 0.5 : 0.25(D) 0 : 0.25 : 0.75

Ans. (C)

112. Match the plants in Column I with their features

listed in the Column II, III & IV.

Column I Column II Column III Column IVTypes of plants

Types of Photosynthesis

Site of Calvincycle

Time of stomata opening

Rice CAM Mesophyll Day

Pineapple C4 Bundle Sheath

Night

Sugarcane C3

Choose the Correct combination.

(A) Rice-C3-Mesophyll-Day, Pineapple-CAM-

Mesophyll-Night, Sugarcane-C4-Bundle

sheath-day

(B) Rice-C3-Mesophyll-Day, Pineapple-CAM-

Mesophyll-Night, Sugarcane-C4-

Mesophyll-Day

(C) Rice-C4-Mesophyll-Day, Pineapple-C3-

Bundlle sheath-Night, Sugarcane-CAM-

Bundle sheath-Day

(D) Rice-CAM-Mesophyll-Day, Pineapple-

CAM-Mesophyll-Day, Sugarcane-C4-

Bundle sheath-Day

Ans. (A)

113. A bacteriophage T2 particle contains within its

head a double-stranded B-from DNA of

molecular weight 1.2 × 108 Da. Assume that

the head of a T2 Phage particle is of 210 nm in

length and the average molecular weight of a

nucleotide is 330 Da. The length of the T2

genome is in the range of

(A) 6 × 105 to 6.4 × 105 nm

(B) 40 × 104 to 41 × 104 nm

(C) 1.8 × 105 to 2 × 105 nm

(D) 6 × 104 to 6.4 × 104 nm

Ans. (D)

ALLEN


30

ALLEN114. In the graph below, where N is population size

and t is time, M represents

dN/dt

NM

(A) specific growth rate.(B) Median population size.(C) carrying capacity.(D)minimum population size without going

extinct.

Ans. (C)115. Match the metabolic pathways in Column I

with their corresponding intermediate moleculeslisted in Column II

Column I Column IIP Krebs cycle i Dihydroxy acetone

phosphateQ Glycolysis ii SuccinateR Electron transport

chainiii Cytochrome c

S Nitrogen fixation iv Glutamatev Glyoxylate

Choose the CORRECT combination.

(A) P-ii, Q-i, R-iii, S-iv

(B) P-i, Q-v, R-iv, S-ii

(C) P-v, Q-i, R-iii, S-iv

(D) P-ii, Q-i, R-iii, S-v

Ans. (A)116. By comparing mitosis and meiosis occurring in

the same organism, which ONE of the followingoptions is CORRECT regarding the DNAcontent per cell ?

(A) Mitotic anaphase > Meiotic anaphase I =Meiotic anaphase II

(B) Mitotic anaphase = Meiotic anaphase I >Meiotic anaphase II

(C) Mitotic anaphase < Meiotic anaphase I =Meiotic anaphase II

(D) Mitotic anaphase = Meiotic anaphase I <

Meiotic anaphase II

Ans. (B)

117. Which ONE of the following is likely to occurupon heating a solution of eukaryotic proteinfrom 20°C to 95°C ?(A) Breakage of disulphide bonds(B) Change in primary structure(C) Hydrolysis of peptide bonds(D) Change in tertiary structure

Ans. (D)118. Which ONE of the following statements is

INCORRECT about the hexokinase-catalysedreaction given below ?Glucose + ATP ® Glucose-6-phosphate+ADP(A) This reaction takes place in the cytoplasm(B) This is an endergonic reaction(C) Folding of hexokinase to fit around the

glucose molecule excludes water from theactive site

(D) This reaction involves an induced fitmechanism in hexokinase

Ans. (B)119. An ecologist samples trees in multiple forest

plots to determine species richness.Which ONE of the following can help determinethe adequacy of sampling effort ?(A) Graph the number of new tree species in

each successive sampling plot.(B) Graph the total number of tree species per

total area for all plots combined.(C) Graph the number of individuals per tree

species in each successive sampling plot.(D)30 sampling plots are sufficient, irrespective

of the forest area.

Ans. (A)120. In medical diagnostics for a disease, sensitivity

(denoted a) of a test refers to the probability thata test result is positive for a person with thedisease whereas specificity (denoted b) refersto the probability that a person without thedisease test negative. A diagnostic test forinfluenza has the values of a = 0.9 and b = 0.9.Assume that the prevalence of influenza in apopulation in 50%. If a randomly chosen persontests negative, what is the probability that theperson actually has influenza ?(A) 0.01 (B) 0.02(C) 0.05 (D) 0.10

Ans. (D)

ALLEN

Documents

ALLEN KVPY -2020 / Srea-SB/SX KVPY – 2020 STREAM - SB/SX